Solving Trig functions?

**dondonlouie** · March 27th 2011, 02:53 PM

I'm not sure how to solve these trig functions;

1. 2sin(theta)−sqrt(2)tan(theta)= 0 ; 0 ≤ theta < 2pi

2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:

**e^(i*pi)** · March 27th 2011, 03:06 PM

1. Multiply through by $]\cos(\theta)$ and solve by factoring

**topsquark** · March 27th 2011, 04:09 PM

Originally Posted by dondonlouie

2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:

It's probably better to rewrite that first line as
$sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan

**dondonlouie** · March 27th 2011, 05:04 PM

Originally Posted by topsquark

It's probably better to rewrite that first line as
$sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan

wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?

**topsquark** · March 27th 2011, 05:19 PM

Originally Posted by dondonlouie

wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?

Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$(sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan

**dondonlouie** · March 27th 2011, 08:18 PM

Originally Posted by topsquark

Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$(sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan

how did you get from sin(theta)(cos(theta)-1)-(cos(theta)-1)= 0
to (sin(theta)-1)(cos(theta)-1)?

**Prove It** · March 27th 2011, 08:35 PM

Take out $\displaystyle (\cos{\theta} - 1)$ as a factor...

**dondonlouie** · March 27th 2011, 09:14 PM

Originally Posted by Prove It

Take out $\displaystyle (\cos{\theta} - 1)$ as a factor...

ohhh okay i see it

thank you so much for helping me

**topsquark** · March 28th 2011, 01:07 PM

Originally Posted by dondonlouie

ohhh okay i see it

My apologies. I thought you already had factored that.

-Dan

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